Question

Almost all employees working for financial companies in New York City receive large bonuses at the end of the year. A sample of 65 employees selected from financial companies in New York City showed that they received an average bonus of $51,000 last year with a standard deviation of $18,000. Construct a 99% confidence interval for the average bonus that all employees working for financial companies in New York City received last year. Round your answers to cents.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $51000

Population standard deviation =    = $18000
Sample size = n =65

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

Margin of error = E = Z/2* ( /n)

= 2.576 * (18000 / 65)

= 5751.2426

At 99% confidence interval estimate of the population mean is,

- E < < + E

51000- 5751.2426< <51000+ 5751.2426

45248.7574< < 56751.2426

(45248.7574,56751.2426)